二项式定理：
(
a
+
b
)
n
=
C
n
0
a
n
+
C
n
1
a
n
?
1
b
+
C
n
2
a
n
?
2
b
2
+
?
+
C
n
n
?
1
a
b
n
?
1
+
C
n
n
b
n
=
∑
i
=
0
n
C
n
i
a
n
?
i
b
i
(a + b)^n = C_n^0a^n + C_n^1a^{n-1}b + C_n^2a^{n-2}b^2 + \cdots + C_n^{n-1}ab^{n-1} + C_n^nb^n = \sum_{i=0}^{n}C_n^i a^{n-i}b^i
(a+b)n=Cn0?an+Cn1?an?1b+Cn2?an?2b2+?+Cnn?1?abn?1+Cnn?bn=i=0∑n?Cni?an?ibi
Stolz 定理：
两个数列
{
x
n
}
\lbrace x_n \rbrace
{xn?} 和
{
y
n
}
\lbrace y_n \rbrace
{yn?}，其中
y
n
y_n
yn? 是无穷大量。如果：
lim
?
n
→
∞
x
n
?
x
n
?
1
y
n
?
y
n
?
1
=
a
\lim_{n\to \infty}\frac{x_n - x_{n-1}}{y_n-y_{n-1}} = a
n→∞lim?yn??yn?1?xn??xn?1??=a
则：
lim
?
n
→
∞
x
n
y
n
=
a
\lim_{n\to \infty} \frac{x_n}{y_n} = a
n→∞lim?yn?xn??=a
夹逼定理：
三个数列
{
x
n
}
\lbrace x_n \rbrace
{xn?} ,
{
y
n
}
\lbrace y_n \rbrace
{yn?} 和
{
z
n
}
\lbrace z_n \rbrace
{zn?}。若：
x
n
≤
y
n
≤
z
n
且
lim
?
n
→
∞
x
n
=
lim
?
n
→
∞
z
n
=
a
x_n \leq y_n \leq z_n \qquad 且 \qquad \lim_{n\to \infty}x_n = \lim_{n\to \infty}z_n = a
xn?≤yn?≤zn?且n→∞lim?xn?=n→∞lim?zn?=a
则：
lim
?
n
→
∞
y
n
=
a
\lim_{n\to \infty}y_n = a
n→∞lim?yn?=a
题(们)
（1）
lim
?
n
→
∞
(
1
+
1
2
+
1
3
+
?
+
1
n
)
1
n
≤
lim
?
n
→
∞
n
1
n
=
1
\lim_{n\to \infty}(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n})^{\frac{1}{n}} \leq \lim_{n\to \infty}n^{\frac{1}{n}} = 1
n→∞lim?(1+21?+31?+?+n1?)n1?≤n→∞lim?nn1?=1
lim
?
n
→
∞
(
1
+
1
2
+
1
3
+
?
+
1
n
)
1
n
≥
lim
?
n
→
∞
1
1
n
=
1
\lim_{n\to \infty}(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n})^{\frac{1}{n}} \geq \lim_{n\to \infty}1^{\frac{1}{n}} = 1
n→∞lim?(1+21?+31?+?+n1?)n1?≥n→∞lim?1n1?=1
∴
lim
?
n
→
∞
(
1
+
1
2
+
1
3
+
?
+
1
n
)
1
n
=
1
\therefore \lim_{n\to \infty}(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n})^{\frac{1}{n}} = 1
∴n→∞lim?(1+21?+31?+?+n1?)n1?=1
（2）
lim
?
n
→
∞
(
1
n
+
1
+
1
n
+
2
+
?
+
1
n
+
n
)
≤
lim
?
n
→
∞
n
n
+
n
=
lim
?
n
→
∞
1
1
+
1
n
=
1
\lim_{n\to \infty}(\frac{1}{n + \sqrt{1}} + \frac{1}{n + \sqrt{2}} + \cdots +\frac{1}{n + \sqrt{n}}) \leq \lim_{n\to \infty} \frac{n}{n + \sqrt{n}} = \lim_{n\to \infty}\frac{1}{1+\frac{1}{\sqrt{n}}} = 1
n→∞lim?(n+1
?1?+n+2
?1?+?+n+n
?1?)≤n→∞lim?n+n
?n?=n→∞lim?1+n
?1?1?=1
lim
?
n
→
∞
(
n
n
+
1
+
n
n
+
2
+
?
+
n
n
+
n
)
≥
lim
?
n
→
∞
n
n
+
1
=
1
1
+
1
n
=
1
\lim_{n\to \infty}({\frac{n}{n +\sqrt{1}} + \frac{n}{n +\sqrt{2}} +\cdots+ \frac{n}{n +\sqrt{n}}}) \geq \lim_{n\to \infty}\frac{n}{n+1} = \frac{1}{1+\frac{1}{n}} = 1
n→∞lim?(n+1
?n?+n+2
?n?+?+n+n
?n?)≥n→∞lim?n+1n?=1+n1?1?=1
∴
lim
?
n
→
∞
(
1
n
+
1
+
1
n
+
2
+
?
+
1
n
+
n
)
=
1
\therefore \lim_{n\to \infty}(\frac{1}{n+\sqrt{1}} + \frac{1}{n +\sqrt{2}} + \cdots + \frac{1}{n +\sqrt{n}}) = 1
∴n→∞lim?(n+1
?1?+n+2
?1?+?+n+n
?1?)=1
（3）
lim
?
n
→
∞
∑
k
=
n
2
(
n
+
1
)
2
1
k
≤
lim
?
n
→
∞
(
n
+
1
)
2
?
n
2
+
1
n
+
1
=
lim
?
n
→
∞
2
=
2
\lim_{n\to \infty}\sum_{k = n^2}^{(n+1)^2}\frac{1}{\sqrt{k}} \leq \lim_{n\to \infty}\frac{(n+1)^2-n^2+1}{n+1} = \lim_{n\to \infty}2 = 2
n→∞lim?k=n2∑(n+1)2?k
?1?≤n→∞lim?n+1(n+1)2?n2+1?=n→∞lim?2=2
lim
?
n
→
∞
∑
k
=
n
2
(
n
+
1
)
2
1
k
≥
(
n
+
1
)
2
?
n
2
+
1
n
=
2
lim
?
n
→
∞
n
n
+
1
=
2
lim
?
n
→
∞
1
1
+
1
n
=
2
\lim_{n\to \infty}\sum_{k = n^2}^{(n+1)^2}\frac{1}{\sqrt{k}} \geq \frac{(n+1)^2-n^2+1}{n} = 2\lim_{n\to \infty}\frac{n}{n+1} = 2\lim_{n\to \infty}\frac{1}{1+\frac{1}{n}} = 2
n→∞lim?k=n2∑(n+1)2?k
?1?≥n(n+1)2?n2+1?=2n→∞lim?n+1n?=2n→∞lim?1+n1?1?=2
（4）
lim
?
n
→
∞
1
?
3
?
5
?
(
2
n
?
1
)
2
?
4
?
6
?
(
2
n
)
≥
lim
?
n
→
∞
(
1
2
)
n
=
0
\lim_{n\to \infty}\frac{1\cdot 3\cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} \geq \lim_{n\to \infty}(\frac{1}{2})^n = 0
n→∞lim?2?4?6?(2n)1?3?5?(2n?1)?≥n→∞lim?(21?)n=0
令
G
=
∏
i
=
1
n
2
i
?
1
2
i
，
P
=
∏
i
=
1
n
2
i
2
i
+
1
令 G = \prod_{i=1}^{n}\frac{2i-1}{2i}，P = \prod_{i=1}^{n}\frac{2i}{2i+1}
令G=i=1∏n?2i2i?1?，P=i=1∏n?2i+12i?
G
<
P
且
G
P
=
1
2
n
+
1
∴
G
<
1
2
n
+
1
→
lim
?
n
→
∞
G
≤
lim
?
n
→
∞
1
2
n
+
1
=
0
G < P 且 GP = \frac{1}{2n+1} \therefore G < \frac{1}{\sqrt{2n+1}} \rightarrow \lim_{n\to \infty}G \leq \lim_{n\to \infty}\frac{1}{\sqrt{2n+1}} = 0
G<P且GP=2n+11?∴G<2n+1
?1?→n→∞lim?G≤n→∞lim?2n+1
?1?=0
lim
?
n
→
∞
1
?
3
?
5
?
(
2
n
?
1
)
2
?
4
?
6
?
(
2
n
)
=
0
\lim_{n\to \infty}\frac{1\cdot 3\cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} = 0
n→∞lim?2?4?6?(2n)1?3?5?(2n?1)?=0
（1）
lim
?
n
→
∞
3
n
2
+
4
n
?
1
n
2
+
1
=
3
+
lim
?
n
→
∞
n
?
1
n
2
+
1
=
3
+
lim
?
n
→
∞
1
?
1
n
n
+
1
n
=
3
\lim_{n\to \infty}\frac{3n^2+4n-1}{n^2+1} = 3 + \lim_{n\to \infty}\frac{n-1}{n^2+1} = 3 + \lim_{n\to \infty}\frac{1-\frac{1}{n}}{n+\frac{1}{n}} = 3
n→∞lim?n2+13n2+4n?1?=3+n→∞lim?n2+1n?1?=3+n→∞lim?n+n1?1?n1??=3
（2）
lim
?
n
→
∞
n
3
+
2
n
2
?
3
n
+
1
2
n
3
?
n
+
3
=
lim
?
n
→
∞
1
+
2
n
+
3
n
2
+
1
n
3
2
?
1
n
+
1
n
3
=
1
2
\lim_{n\to \infty}\frac{n^3+2n^2-3n+1}{2n^3-n+3} = \lim_{n\to \infty}\frac{1 + \frac{2}{n} + \frac{3}{n^2} + \frac{1}{n^3}}{2 - \frac{1}{n} + \frac{1}{n^3}} = \frac{1}{2}
n→∞lim?2n3?n+3n3+2n2?3n+1?=n→∞lim?2?n1?+n31?1+n2?+n23?+n31??=21?
（3）
lim
?
n
→
∞
3
n
+
n
3
3
n
+
1
+
(
n
+
1
)
3
=
lim
?
n
→
∞
1
+
n
3
3
n
3
+
(
n
+
1
)
3
3
n
=
1
3
（
lim
?
n
→
∞
n
k
a
n
=
0
证
明
见
8
）
\lim_{n\to \infty}\frac{3^n + n^3}{3^{n+1}+(n+1)^3} = \lim_{n\to \infty}\frac{1+\frac{n^3}{3^n}}{3 + \frac{(n+1)^3}{3^n}} = \frac{1}{3} （\lim_{n\to \infty}\frac{n^k}{a^n} = 0 证明见 8）
n→∞lim?3n+1+(n+1)33n+n3?=n→∞lim?3+3n(n+1)3?1+3nn3??=31?（n→∞lim?annk?=0证明见8）
（4）
lim
?
n
→
∞
(
n
2
+
1
n
?
1
)
s
i
n
n
π
2
=
[
(
lim
?
n
→
∞
n
2
+
1
n
)
?
1
]
lim
?
n
→
∞
s
i
n
n
π
2
=
0
\lim_{n\to \infty}(\sqrt[n]{n^2+1}-1)sin\frac{n\pi}{2} = \Bigg[\Big(\lim_{n\to \infty}\sqrt[n]{n^2+1}\Big)-1\Bigg]\lim_{n\to \infty}sin\frac{n\pi}{2} = 0
n→∞lim?(nn2+1
??1)sin2nπ?=[(n→∞lim?nn2+1
?)?1]n→∞lim?sin2nπ?=0
（5）
lim
?
n
→
∞
n
(
n
+
1
?
n
)
=
lim
?
n
→
∞
(
n
2
+
n
?
n
)
(
n
2
+
n
+
n
)
n
2
+
n
+
n
=
lim
?
n
→
∞
n
n
2
+
n
+
n
=
lim
?
n
→
∞
1
1
+
1
+
1
n
=
1
2
\begin{aligned} &\lim_{n\to \infty}\sqrt{n}(\sqrt{n+1} - \sqrt{n}) = \lim_{n\to \infty}\frac{(\sqrt{n^2+n}-n)(\sqrt{n^2+n} + n)}{\sqrt{n^2+n}+n}\\ = &\lim_{n\to \infty}\frac{n}{\sqrt{n^2+n}+n} = \lim_{n\to \infty}\frac{1}{1 + \sqrt{1+\frac{1}{n}}} = \frac{1}{2} \end{aligned}
=?n→∞lim?n
?(n+1
??n
?)=n→∞lim?n2+n
?+n(n2+n
??n)(n2+n
?+n)?n→∞lim?n2+n
?+nn?=n→∞lim?1+1+n1?
?1?=21??
（6）
lim
?
n
→
∞
(
1
?
1
2
2
)
(
1
?
1
3
2
)
?
(
1
?
1
n
2
)
=
lim
?
n
→
∞
2
2
?
1
2
2
?
3
2
?
1
3
2
?
4
2
?
1
4
2
?
n
2
?
1
n
2
=
lim
?
n
→
∞
(
2
+
1
)
(
2
?
1
)
(
3
+
1
)
(
3
?
1
)
(
4
+
1
)
(
4
?
1
)
?
(
n
+
1
)
(
n
?
1
)
2
2
?
3
2
?
4
2
?
n
2
=
lim
?
n
→
∞
n
+
1
2
n
=
lim
?
n
→
∞
1
+
1
n
2
=
1
2
\begin{aligned} &\lim_{n\to \infty} (1-\frac{1}{2^2})(1-\frac{1}{3^2})\cdots (1-\frac{1}{n^2})\\ = & \lim_{n\to \infty}\frac{2^2-1}{2^2} \cdot \frac{3^2-1}{3^2} \cdot \frac{4^2-1}{4^2} \cdots \frac{n^2-1}{n^2}\\ = &\lim_{n\to \infty}\frac{(2+1)(2-1)(3+1)(3-1)(4+1)(4-1) \cdots (n+1)(n-1)}{2^2\cdot3^2\cdot4^2\cdots n^2} \\ = &\lim_{n\to \infty}\frac{n+1}{2n} = \lim_{n\to \infty}\frac{1+\frac{1}{n}}{2} = \frac{1}{2} \end{aligned}
===?n→∞lim?(1?221?)(1?321?)?(1?n21?)n→∞lim?2222?1??3232?1??4242?1??n2n2?1?n→∞lim?22?32?42?n2(2+1)(2?1)(3+1)(3?1)(4+1)(4?1)?(n+1)(n?1)?n→∞lim?2nn+1?=n→∞lim?21+n1??=21??
（7）
lim
?
n
→
∞
n
(
n
2
+
1
4
?
n
+
1
)
=
lim
?
n
→
∞
n
n
2
+
1
4
+
n
2
+
2
n
+
1
4
=
lim
?
n
→
∞
n
n
2
+
1
?
n
2
+
2
n
+
1
n
2
+
1
4
+
n
2
+
2
n
+
1
4
=
lim
?
n
→
∞
n
?
2
n
(
n
2
+
1
4
+
n
2
+
2
n
+
1
4
)
(
n
2
+
1
+
n
2
+
2
n
+
1
)
=
lim
?
n
→
∞
?
2
n
3
2
(
n
2
+
1
4
+
n
2
+
2
n
+
1
4
)
(
n
2
+
1
+
n
2
+
2
n
+
1
)
=
lim
?
n
→
∞
?
2
n
3
2
4
n
=
?
1
2
\begin{aligned} & \lim_{n\to \infty}\sqrt{n}(\sqrt[4]{n^2+1}- \sqrt{n+1})\\ = & \lim_{n\to \infty}\sqrt{n}{\sqrt[4]{n^2+1} + \sqrt[4]{n^2+2n+1}}\\ = & \lim_{n\to \infty}\sqrt{n}\frac{\sqrt{n^2+1} - \sqrt{n^2+2n+1}}{\sqrt[4]{n^2+1}+\sqrt[4]{n^2+2n+1}}\\ = & \lim_{n\to \infty}\sqrt{n}\frac{-2n}{\left( \sqrt[4]{n^2+1}+\sqrt[4]{n^2+2n+1} \right) \left( \sqrt{n^2+1}+\sqrt{n^2+2n+1} \right) }\\ = & \lim_{n\to \infty} \frac{-2n^{\frac{3}{2}}}{\left( \sqrt[4]{n^2+1}+\sqrt[4]{n^2+2n+1} \right) \left( \sqrt{n^2+1}+\sqrt{n^2+2n+1} \right) }\\ = & \lim_{n\to \infty}\frac{-2n^{\frac{3}{2}}}{4n} = -\frac 12 \end{aligned}
=====?n→∞lim?n
?(4n2+1
??n+1
?)n→∞lim?n
?4n2+1
?+4n2+2n+1
?n→∞lim?n
?4n2+1
?+4n2+2n+1
?n2+1
??n2+2n+1
??n→∞lim?n
?(4n2+1
?+4n2+2n+1
?)(n2+1
?+n2+2n+1
?)?2n?n→∞lim?(4n2+1
?+4n2+2n+1
?)(n2+1
?+n2+2n+1
?)?2n23??n→∞lim?4n?2n23??=?21??
（8）
lim
?
n
→
∞
(
1
2
+
3
2
2
+
?
+
2
n
?
1
2
n
)
\lim_{n\to \infty}(\frac{1}{2} + \frac{3}{2^2} + \cdots + \frac{2n-1}{2^n})
n→∞lim?(21?+223?+?+2n2n?1?)
- 证明：若
lim
?
n
→
∞
a
n
=
a
\lim_{n\to \infty} a_n = a
limn→∞?an?=a 则
lim
?
n
→
∞
a
1
+
a
2
+
a
3
+
?
+
a
n
n
=
a
\lim_{n\to \infty}\frac{a_1+a_2+a_3+\cdots + a_n}{n} = a
limn→∞?na1?+a2?+a3?+?+an??=a
- 设
a
n
>
0
a_n > 0
an?>0， 且
lim
?
n
→
∞
a
n
=
a
\lim_{n\to \infty}a_n = a
limn→∞?an?=a，证明：
lim
?
n
→
∞
a
1
a
2
?
a
n
n
=
a
\lim_{n\to \infty}\sqrt[n]{a_1a_2\cdots a_n} = a
limn→∞?na1?a2??an?
?=a
- 若
a
n
>
0
a_n > 0
an?>0 （n = 1， 2， 3…），且
lim
?
n
→
∞
a
n
n
=
a
\lim_{n\to \infty}\sqrt[n]{a_n} = a
limn→∞?nan?
?=a
- 求
lim
?
n
→
∞
1
2
+
3
3
+
5
2
+
?
+
(
2
n
+
1
)
2
n
3
\lim_{n\to \infty}\frac{1^2 + 3^3 + 5^2 + \cdots + (2n+1)^2}{n^3}
limn→∞?n312+33+52+?+(2n+1)2?
- 用 Stolz 定理求
lim
?
n
→
∞
l
o
g
a
n
n
\lim_{n\to \infty}\frac{log_an}{n}
limn→∞?nloga?n? 和
lim
?
n
→
∞
n
k
a
n
\lim_{n\to \infty}\frac{n^k}{a^n}
limn→∞?annk?
- 设
{
x
n
}
\lbrace x_n \rbrace
{xn?} 是无穷大量，
∣
y
n
∣
≥
δ
>
0
|y_n| \geq \delta > 0
∣yn?∣≥δ>0，证明
{
x
n
y
n
}
\lbrace x_ny_n \rbrace
{xn?yn?} 是无穷大量。
|